Let $\phi$ be a ring homomorphism $\mathbb{C}[x,y,z] \to \mathbb{C}[s,t]$ such that $\phi(x) = s,\ \phi(y) = st,$ and $\phi(z) = t^2.$ Let $R = \operatorname{Im}\phi.$ Let $(a,b,c)\in\mathbb{C}^3$ and suppose $\mathfrak{m} = (s-a,st-b,t^2-c)$ is a maximal ideal of $R.$ The question asks me to find all $(a,b,c) \in \mathbb{C}^3$ such that the ideal $\mathfrak{m}R_\mathfrak{m}$ is not generated by two elements.
These have been my ideas so far:
First of all, by the first isomorphism theorem, $$R \cong \mathbb{C}[x,y,z]/(y^2-x^2z).$$ The ideal $\mathfrak{m}$ then corresponds to the ideal $(x-a,y-b,z-c)$ in $\mathbb{C}[x,y,z].$
Certainly, $\mathfrak{m}R_\mathfrak{m}$ is generated by at most $3$ elements, as the images of the generators of $\mathfrak{m}$ will generate $\mathfrak{m}R_\mathfrak{m}.$ So we need to find $(a,b,c)$ such that $\mathfrak{m}R_\mathfrak{m}$ is principal or generated by three elements.
By Krull's height theorem, if $\mathrm{ht}(\mathfrak{m}R_\mathfrak{m}) \geq 2,$ then $\mathfrak{m}R_\mathfrak{m}$ cannot be principal, as the height of a proper ideal is bounded above by the number of generators of that ideal. Furthermore, $\mathrm{ht}(\mathfrak{m}R_\mathfrak{m}) = \mathrm{ht}(\mathfrak{m})$ as the localisation gives us a bijection between prime ideals of $R$ contained in $\mathfrak{m}$ and prime ideals of $R_\mathfrak{m}.$
As commented above, $\mathfrak{m}$ corresponds to the ideal $(x-a,y-b,z-c)$ in $\mathbb{C}[x,y,z],$ so we need to consider the prime ideals $\mathfrak{p} \subset \mathbb{C}[x,y,z]$ such that $(y^2-x^2z) \subseteq \mathfrak{p} \subseteq (x-a,y-b,z-c).$
For example, if we consider $(a,b,c) = (0,0,0),$ we would have the following chain of ideals in $\mathbb{C}[x,y,z]:$ $$(y^2-x^2z) \subsetneq (x,y) \subsetneq (x,y,z),$$ so $(x,y,z)$ has height at least $2$, hence the corresponding maximal ideal $\mathfrak{m}R_\mathfrak{m}$ cannot be principal.
However, this method still doesn't allow me to decide whether $(x,y,z)$ has two or three generators in the localisation. Also, for $(a,b,c) \neq (0,0,0),$ I'm not sure how to find the height of the ideal $(s-a,st-b,t^2-c).$
Of course, I'm not even sure if this method is the correct method (as the height only gives the lower bound for the number of generators). If anybody has better ideas or any hints, it'd be much appreciated!
EDIT: I realised I can actually show that all the prime ideals of $\mathbb{C}[s,t]$ are generated by at most $2$ elements, so $\mathfrak{m}$ is generated by at most two elements. So I guess this means we just need to decide whether $\mathfrak{m}R_\mathfrak{m}$ is principal or not?
EDIT2: Actually, a prime ideal in $R$ isn't necessarily prime in $\mathbb{C}[s,t],$ so I guess the above doesn't quite work.
EDIT3: I feel like I have a possible solution. It'd be great if somebody could check them!
Possible solution: Let $d$ be the minimal number of generators of $\mathfrak{n} = \mathfrak{m}R_\mathfrak{m}.$ Then by Nakayama's Lemma, $d$ is also the minimal number of generators of $\mathfrak{n}/\mathfrak{n}^2.$ So $d = \dim_{R/\mathfrak{m}}\mathfrak{n}/\mathfrak{n}^2.$ Note that the image of $\{x-a,y-b,z-c\}$ in $R_\mathfrak{m}$ generate $\mathfrak{n},$ so $\{(x-a)^2,(y-b^2),(z-c)^2,(x-a)(y-b),(y-b)(z-c),(z-c)(x-a)\}$ generate $\mathfrak{n}^2.$ I'm going to try to show that $d=3.$ Suppose $u(x-a)+v(y-b)+w(z-c) \in \mathfrak{n}^2$ for $u,v,w\in R/\mathfrak{m} \cong \mathbb{C}.$ The equation above is linear, so we cannot have the square generators (such as $(x-a)^2$). But every other generator contains $xy$, $yz$, or $zx,$ which isn't possible. So $u=v=w=0.$ So the three generators (modulo $\mathfrak{n}^2$) are linearly independent, hence $d=3$. So no choice of $(a,b,c)$ works.
First of all, let's review what it is the relation of the number of generators of $mR_m$ and the non-singularity.
Denote $\mathfrak{a}=(x-a,y-b,z-c)\subset \mathbb{C}[x,y,z]$ the maximal ideal in the polynomial ring corresponding to the point $P=(a,b,c)\in \mathbb{C}^3$. Then the map $$\psi:\mathfrak{a}/\mathfrak{a}^2\to \mathbb{C}^3$$ given by $$\psi(g)=(\frac{\partial g}{\partial x}(P),\frac{\partial g}{\partial y}(P),\frac{\partial g}{\partial z}(P))$$ is an isomorphism of $\mathbb{C}$-vector spaces. This can be shown for example by an explicit computation in the case $a=b=c=0$, and one can reduce to this case.
Now, for a given ideal say $I=(f)$, for $f\in \mathbb{C}[x,y,z]$, the subspace generated by the image $\psi(f)$ is isomorhic to $(I+\mathfrak{a}^2)/\mathfrak{a}^2$. Now, if you denote $$\mathfrak{m}=(\overline{x-a},\overline{y-b},\overline{z-c})\subset \mathbb{C}[x,y,z]/I,$$ then one has that $$\mathfrak{a}/(I+\mathfrak{a}^2)\cong \mathfrak{m}/\mathfrak{m}^2,$$ the map induced by the quotient map. Hence you get that $$\operatorname{rank}_{\mathbb{C}}(\mathfrak{m}/\mathfrak{m}^2)=3-\operatorname{rank}_{\mathbb{C}}(\langle \psi(f) \rangle_{\mathbb{C}}).$$ Hence $\operatorname{rank}_{\mathbb{C}}(\mathfrak{m}/\mathfrak{m}^2)=2$ unless $\psi(f)=(0,0,0)$, in which case it is three.
Now, $\psi(f)=(0,0,0)$ if and only if $a=b=0$ and $c$ undetermined. To finish one only needs to recall that $\operatorname{rank}_{\mathbb{C}}(\mathfrak{m}/\mathfrak{m}^2)$ is equal to the minimal number of generators of $\mathfrak{m}$.
Hence the answer of your question is $a=b=0$.
Observe that what I did can be done for any number of variables, and even when the ideal $I$ is not generated by one polynomial, in this case considering the rank of the jacobian matrix. So the number of generators of a maximal ideal $\mathfrak{m}$ is equal to the dimension of the ring if and only if the corresponding point $P$ is non singular.